Base change for non-flat coherent sheaves and affine maps Let $A$ be a finitely generated $k$-algebra, where $k$ is a field, let $I$ be an ideal in $A$, let $M$ be a finitely generated $A/I$-module, and let $M^{\prime}$ denote $M$ considered as an $A$-module. Let $B$ be a finitely generated $A$-algebra. Is it true (perhaps under some additional conditions on $A$ and $I$, though I need the case when $B$ is not flat over $A$) that $M^{\prime} \otimes^L_A B$ and $M \otimes^L_{A/I} B/IB$ are quasi-isomorphic as complexes of $B$-modules? 
 A: There is an associativity identity for total derived tensor products, cf. Stacks Project Tag 08YU.  In your case, for the triple of rings, $$A\twoheadrightarrow A/I \xrightarrow{\text{Id}}A/I,$$ this gives an equivalence in the derived category, $$ M\otimes_{A/I}^{\textbf{L}}(A/I\otimes_A^{\textbf{L}} B) \cong M\otimes_A^{\textbf{L}} B. $$  Thus, if the natural truncation morphism, $$A/I\otimes_A^{\textbf{L}}B \to h_0(A/I\otimes_A^{\textbf{L}}B), \text{ i.e., }\ A/I\otimes_A^{\textbf{L}}B \to B/IB, $$
is an equivalence in the derived category, then that gives the equivalence that you are asking about, $$M\otimes_{A/I}^{\textbf{L}} B/IB \cong M\otimes_A^{\textbf{L}}B.$$
As the OP points out, the natural truncation morphism is an equivalence in the derived category if and only if $h_i(A/I\otimes_A^{\textbf{L}} B)$ vanishes for every $i>0$, i.e., if and only if $\text{Tor}_i^A(A/I,B)$ vanishes for every $i>0$.  The OP asks whether it might suffice to have vanishing of $\text{Tor}_i^A(A/I,M)$ for $i>0$?  It is hard for me to imagine any situation where this vanishing would hold, since $\text{Tor}_1^A(A/I,M)$ is naturally isomorphic to $I\otimes_A M$ for every $A/I$-module $M$.  
